### 3. Rationality or Irrationality of Polynomial Roots

 It is quite interesting to note that any root of a polynomial with integer coefficients and leading coefficient 1 is either an integer or irrational.  A key component of the proof is the Rational Root Theorem, which states that if we have a polynomial with integer coefficients, then any rational root r/s (in simplest form) has r as a factor of the constant term and s as a factor of the leading coefficient.  We can prove it as follows.  Consider the roots of a general polynomial with integer coefficients:anxn +  an - 1xn - 1 + ... + a2x2 + a1x1 + a0 = 0If we substitute a rational root r/s (again, in simplest form) into this equation, and then multiply through by sn-1, we get:anrn/s +  an - 1rn - 1 + ... + a2r2sn-3 + a1rsn-2 + a0sn-1 = 0Because all the coefficients ai (for all i between 0 and n) are integers, we see right away that all terms except the first are integers.  This actually forces the first term anrn/s to be an integer as well, since everything sums to zero.  For this to happen, s must divide anrn; since we assume that r and s are relatively prime, the only possibility is that s divides the leading coefficient an.Now we return to the original equation, substitute r/s for x, and this time multiply through by sn/r, yielding:anrn-1 +  an - 1rn - 2s + an-2rn-3s2 + ...+ a1sn-1 + a0sn/r = 0By the same reasoning, a0sn/r is an integer, so r divides  a0sn/r; again, since r and s are relatively prime, r must divide the constant term a0.  Combining the two parts completes the proof.  (Source:  The Math Forum, Ask Dr. Math, , accessed 13 July 2009)Back to the problem at hand, considering a polynomial with integer coefficients and leading coefficient 1, we see that all rational roots are of the form p/1 or -p/1, by the Rational Root Theorem.  Therefore, all rational roots of this polynomial are integers.  The only other possibility is that of irrational roots.  (If we are given a polynomial with a constant term of 0, we simply factor out the highest power of x possible, and then proceed with the proof.  For example, we would rewrite the polynomial x4 - 5x3 + 2x2 as x2(x2 - 5x + 2) and then consider the x2 - 5x + 2 portion.  The only change is that the original polynomial has a double root of x =  0, which, however, does not interfere with the proof as this root would simply fall into the integer category.)As an example of this occurence, we can consider the case of the polynomial x3 + 2x2 - 10x + 7.  All possible rational roots are 1/1, 7/1, and their opposites; we find that only x = 1 works.  Factoring out (x - 1) from the expression, we get:(x - 1)(x2 + 3x - 7)and applying the quadratic formula to the second-degree portion yields:x = (-3 + √(32 - 4 * 1 * (-7)))/(2 * 1) = (-3 + √37)/2andx = (-3 - √(32 - 4 * 1 * (-7)))/(2 * 1) = (-3 - √37)/2.Since 37 is not a perfect square, √37 is irrational, and so are the roots of the quadratic factor of the polynomial, by the theorems proved in the previous section.  So in the end, we have one integer root, 1, and two irrational roots, (-3 + √37)/2 and (-3 - √37)/2.
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